Hardy-Weinberg equilibrium
The Hardy-Weinberg principle links allele and genotype frequencies in a population. If you know how many people have a recessive disease, you can work out how many are hidden carriers.
Rules
- p + q = 1 (allele frequencies)
- p² + 2pq + q² = 1 (AA, Aa, aa)
- Affected (aa) = q² → q = square root of the affected share
- Carriers (Aa) = 2pq
- Assumptions: large population, random mating, no selection, mutation or migration
Your turn
Three problems on this topic. Type probabilities as a percentage number, e.g. 18.75.
Task
Cystic fibrosis is inherited autosomal recessive. In a population in Hardy-Weinberg equilibrium, 1 in 3,600 people is affected. Work out the allele frequencies q (recessive) and p and the share of heterozygous carriers.
Allele frequency q
Allele frequency p
Share of heterozygous carriers 2pq in %
Probabilities as a percentage number (25 or 18.75), rounded to 2 decimal places; allele frequencies as a decimal (0.02).
Worked example
Albinism is inherited autosomal recessive. In a population in Hardy-Weinberg equilibrium, 1 in 4,900 people is affected. Work out the allele frequencies q (recessive) and p and the share of heterozygous carriers.
Worked solution
- 1.Affected people are aa: q² = 1/4,900 → q = √(1/4,900) = 1/70 = 0.0143
- 2.p = 1 − q = 0.9857
- 3.Carriers: 2pq = 2 × 0.9857 × 0.0143 = 0.0282 = 2.82% (about 1 in 36)
Typical mistake: Taking the share of affected people as q instead of q². Affected people are aa, their share is q × q; q is its square root.
Common questions
Why are there so many carriers of rare diseases?
Because 2pq is about 2q when q is small, while q² is far smaller. With 1 affected person in 2,500, about 1 in 25 is a carrier.